根据特定单词拆分字符串列表答案

【问题标题】：Splitting a list of strings based on specific words根据特定单词拆分字符串列表
【发布时间】：2021-02-13 02:20:21
【问题描述】：

我有一个包含特定文件路径的列表，例如

path_list =['Animals/dog_00238/2D_rendering/131/view/full/0',
          'Animals/dog_00238/2D_rendering/131/view/full/1',
          'Animals/dog_00238/2D_rendering/131/view/full/2',
          'Animals/dog_00238/2D_rendering/906271/view/full/1',
          'Animals/dog_00239/2D_rendering/906271/view/full/2',
          'Animals/dog_00239/2D_rendering/965947/view/full/0',
          'Animals/dog_00239/2D_rendering/965947/view/full/1',
          'Animals/dog_00239/2D_rendering/965947/view/full/2',<------ Last time dog_00239 appears #first break point
          'Animals/dog_00240/2D_rendering/965947/view/full/3',
          'Animals/dog_00240/2D_rendering/982160/view/full/0',
          'Animals/dog_00240/2D_rendering/982160/view/full/1',
          'Animals/dog_00241/2D_rendering/982160/view/full/2',
          'Animals/dog_00241/2D_rendering/141/view/full/0',
          'Animals/dog_00241/2D_rendering/141/view/full/1',<------ Last time dog_00241 appears #Second breakpoint
          'Animals/dog_00242/2D_rendering/1881/view/full/3',
          'Animals/dog_00242/2D_rendering/1881/view/full/4',
          'Animals/dog_00242/2D_rendering/2487/view/full/0',
          'Animals/dog_00242/2D_rendering/2487/view/full/1',]

我想做的是将此列表拆分为 3 个新的较小列表。

第一个列表将从开始到第一个断点

Fisrt_list = ['Animals/dog_00238/2D_rendering/131/view/full/0',
         'Animals/dog_00238/2D_rendering/131/view/full/1',
         'Animals/dog_00238/2D_rendering/131/view/full/2',
         'Animals/dog_00238/2D_rendering/906271/view/full/1',
         'Animals/dog_00239/2D_rendering/906271/view/full/2',
         'Animals/dog_00239/2D_rendering/965947/view/full/0',
         'Animals/dog_00239/2D_rendering/965947/view/full/1',
         'Animals/dog_00239/2D_rendering/965947/view/full/2',]<------ Last time dog_00239 appears #first break point

第二个列表将在第一个断点之后直到第二个断点

second_list = ['Animals/dog_00240/2D_rendering/965947/view/full/3',
     'Animals/dog_00240/2D_rendering/982160/view/full/0',
     'Animals/dog_00240/2D_rendering/982160/view/full/1',
     'Animals/dog_00241/2D_rendering/982160/view/full/2',
     'Animals/dog_00241/2D_rendering/141/view/full/0',
     'Animals/dog_00241/2D_rendering/141/view/full/1',]<------ Last time dog_00241 appears #Second breakpoint

第三个列表将在第二个断点之后直到结束

third_list = ['Animals/dog_00242/2D_rendering/1881/view/full/3',
     'Animals/dog_00242/2D_rendering/1881/view/full/4',
     'Animals/dog_00242/2D_rendering/2487/view/full/0',
     'Animals/dog_00242/2D_rendering/2487/view/full/1',]

到目前为止，我尝试的是，我创建了一个 for 循环，并在满足 if 条件时将其中断，然后查看我得到的结果，例如：

new_paths = []
fisrt_breakpoint = 'dog_00239'

for i in range (len(path_list)):
new_paths.append(path_list[i])
   if fisrt_breakpoint in path_list:
    break

这确实拆分了列表并保留了其中的一小部分，但它保留在 new_paths 中的是我设置的断点之后的元素，而不是该断点之前的元素。但我也无处设置条件，因为我不知道该怎么做，在这个词最后一次出现在这个列表中时用作断点，因为我假设即使上面的代码有效，它也会拆分它第一次看到“dog_00239”这个词。

我也试过了：

new_paths = [x for x in path_list if first_breakpoint in x]

但这会获取所有包含“dog_00239”的项目，而我想要做的是让所有内容都到此为止。

提前感谢您的宝贵时间，我真的很感激任何帮助，对于任何错误，或者我对我的问题不是很清楚，因为我是这里和 Python 的新手。

【问题讨论】：

标签： python list

【解决方案1】：

您可以使用itertools.groupby

import itertools
import re

# this is the function you define ranges
def get_breakpoint_id(x):
    if(x <= 239): # i.e. /dog_00239
        return 0
    elif(239 < x <= 241): # i.e. /dog_00240 to /dog_00241
        return 1
    else:
        return 2


[list(g) for _,g in itertools.groupby(path_list, key=lambda x: get_breakpoint_id(int(re.findall(r'dog_(\d+)/', x)[0])))]

[['Animals/dog_00238/2D_rendering/131/view/full/0',
  'Animals/dog_00238/2D_rendering/131/view/full/1',
  'Animals/dog_00238/2D_rendering/131/view/full/2',
  'Animals/dog_00238/2D_rendering/906271/view/full/1',
  'Animals/dog_00239/2D_rendering/906271/view/full/2',
  'Animals/dog_00239/2D_rendering/965947/view/full/0',
  'Animals/dog_00239/2D_rendering/965947/view/full/1',
  'Animals/dog_00239/2D_rendering/965947/view/full/2'],
 ['Animals/dog_00240/2D_rendering/965947/view/full/3',
  'Animals/dog_00240/2D_rendering/982160/view/full/0',
  'Animals/dog_00240/2D_rendering/982160/view/full/1',
  'Animals/dog_00241/2D_rendering/982160/view/full/2',
  'Animals/dog_00241/2D_rendering/141/view/full/0',
  'Animals/dog_00241/2D_rendering/141/view/full/1'],
 ['Animals/dog_00242/2D_rendering/1881/view/full/3',
  'Animals/dog_00242/2D_rendering/1881/view/full/4',
  'Animals/dog_00242/2D_rendering/2487/view/full/0',
  'Animals/dog_00242/2D_rendering/2487/view/full/1']]

【讨论】：

刚刚尝试了您的解决方案，它确实按预期工作，但我根本没有使用 itertools.groupby，所以我想问你一些问题。首先，新列表存储在哪里，之后我如何访问它们并对它们执行其他操作？还有在..(int(re.findall(r'dog_(\d+)/', x)[0])))] 在 _(\d+)/' 之后这是如何工作的，直到现在我明白了，但是 ,x ?。再次感谢您
您可以将输出存储在新列表output = [list(g) for _,g in itertools.groupby(path_list, key=lambda x: get_breakpoint_id(int(re.findall(r'dog_(\d+)/', x)[0])))] 中。然后你可以使用output 列表。
(int(re.findall(r'dog_(\d+)/', x)[0])))这里其实regular expression是用来提取数字的dog_XXXX。然后根据提取到的XXXX进行分组。
您可以在线阅读有关itertools.groupby() 的更多信息。 Basicall id有一个列表[1,1,1,2,2,1,1,4,4,4]它可以像(1,1,1), (2,2), (1,1), (4,4,4)一样分组它

【解决方案2】：

对于不涉及导入包的替代解决方案，您可以使用嵌套的 for 循环：

breakpoints = ["dog_00240", "dog_00242"] 
path_list =['Animals/dog_00238/2D_rendering/131/view/full/0',
          'Animals/dog_00238/2D_rendering/131/view/full/1',
          'Animals/dog_00238/2D_rendering/131/view/full/2',
          'Animals/dog_00238/2D_rendering/906271/view/full/1',
          'Animals/dog_00239/2D_rendering/906271/view/full/2',
          'Animals/dog_00239/2D_rendering/965947/view/full/0',
          'Animals/dog_00239/2D_rendering/965947/view/full/1',
          'Animals/dog_00239/2D_rendering/965947/view/full/2',
          'Animals/dog_00240/2D_rendering/965947/view/full/3',
          'Animals/dog_00240/2D_rendering/982160/view/full/0',
          'Animals/dog_00240/2D_rendering/982160/view/full/1',
          'Animals/dog_00241/2D_rendering/982160/view/full/2',
          'Animals/dog_00241/2D_rendering/141/view/full/0',
          'Animals/dog_00241/2D_rendering/141/view/full/1',
          'Animals/dog_00242/2D_rendering/1881/view/full/3',
          'Animals/dog_00242/2D_rendering/1881/view/full/4',
          'Animals/dog_00242/2D_rendering/2487/view/full/0',
          'Animals/dog_00242/2D_rendering/2487/view/full/1',]
lists = []
new_list = []

for file_path in path_list:
    for point in breakpoints:
        # If point is a breakpoint, add new_list to lists
        if point in file_path:
            lists.append(new_list)
            new_list = []
            # Remove breakpoint
            breakpoints.remove(point)
            continue
    # Otherwise, add path to new_list
    new_list.append(file_path)
# Add final new_list to lists
lists.append(new_list)
        
first = lists[0]
second = lists[1]
third = lists[2]

print(first, second, third)
# Prints 
""" (['Animals/dog_00238/2D_rendering/131/view/full/0',  
 'Animals/dog_00238/2D_rendering/131/view/full/1', 
'Animals/dog_00238/2D_rendering/131/view/full/2', 
'Animals/dog_00238/2D_rendering/906271/view/full/1', 
'Animals/dog_00239/2D_rendering/906271/view/full/2', 
'Animals/dog_00239/2D_rendering/965947/view/full/0', 
'Animals/dog_00239/2D_rendering/965947/view/full/1', 
'Animals/dog_00239/2D_rendering/965947/view/full/2'], 
['Animals/dog_00240/2D_rendering/965947/view/full/3', 
'Animals/dog_00240/2D_rendering/982160/view/full/0', 
'Animals/dog_00240/2D_rendering/982160/view/full/1', 
'Animals/dog_00241/2D_rendering/982160/view/full/2', 
'Animals/dog_00241/2D_rendering/141/view/full/0', 
'Animals/dog_00241/2D_rendering/141/view/full/1'], 
['Animals/dog_00242/2D_rendering/1881/view/full/3', 
'Animals/dog_00242/2D_rendering/1881/view/full/4', 
'Animals/dog_00242/2D_rendering/2487/view/full/0', 
'Animals/dog_00242/2D_rendering/2487/view/full/1']) """

请注意，我将断点设置为要中断的第一个路径，而不是到达它的最后一个实例后要中断的路径。

【讨论】：

刚刚尝试了您提出的解决方案，它确实工作得很好，我的断点是总体起点 (0) 和总体终点 (10000) 的终点，因为我有成千上万条这样的路径，我恐怕在那里可能是跳跃，有时下一个起点可能在数字上不一定是下一个，即。 239 接下来 240 但我可以找到确切的下一个起点并按照您的建议。

【解决方案3】：

简单紧凑的解决方案：

breakpoints_numbers = [0,239,241,999]
breakpoints_paths = [f'Animals/dog_00{b}' for b in breakpoints_numbers]
pathslist = [sublist for sublist in [[p for p in path_list if p>b[0] and p<=b[1]] for b in zip(breakpoints_paths[:-1],breakpoints_paths[1:])] if sublist != []]

在第一行创建一个断点列表，包括 000 和 999。

在第二行中，它们被转换为路径，可用于比较。

在第三行中，您有您想要的列表列表，这些列表仅在下限进行过滤，并在上限进行过滤，这正是您所期望的。

【讨论】：

我刚刚尝试了您的解决方案，但您提出的确切示例并给出了结果 [['Animals/dog_00238/2D_rendering/131/view/full/0', 'Animals/dog_00238/2D_rendering/131 /view/full/1'、'Animals/dog_00238/2D_rendering/131/view/full/2'、'Animals/dog_00238/2D_rendering/906271/view/full/1'、'Animals/dog_00239/2D_rendering/906271/view /full/2', 'Animals/dog_00242/2D_rendering/1881/view/full/3', 'Animals/dog_00242/2D_rendering/1881/view/full/4', 'Animals/dog_00242/2D_rendering/2487/view/full /0', 'Animals/dog_00242/2D_rendering/2487/view/full/1'], [], []]
所以它创建了一个包含所有元素的新列表，最后两个为空
你是对的，还有空的子列表。我已经用清理更新了答案以避免它们。该版本正在添加[sublist for sublist in __dirypathslistwithemptysublists__ if sublist != []]
它不仅会删除最后的空文件，还会删除中间的空文件，一旦它们看起来相当不必要 - 至少对我来说，我不知道。
感谢您的更新，但 pathslist = [sublist for sublist in [[p for p in path_list if p>b[0] and p<=b[1]] for b in zip(breakpoints_paths[:-1],breakpoints_paths[1:]) if sublist != []]] 。我认为最后缺少 ] ，除此之外，我认为您的意思是 sublist!= [] 而不是 sublist !+ [] 。但即使进行了这些更改，它也会引发NameError: name 'sublist' is not defined。